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I am studying the series and parallel combinations of capacitors and resistors. I do know that voltage is divided in series whereas it's the same for all circuit components in parallel combination and the vice versa for current, I just want to know the logic behind it. I have read other posts about it, but I don't quite understand them so I'd just like a summarized, to the main point answer. Any help would be appreciated.

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  • $\begingroup$ There are MANY questions posted on this topic addressing this in many contexts and many ways. Don't say "I don't understand" without being specific regarding what you don't understand after you have read them AND the links within them. $\endgroup$
    – Bill N
    Commented Apr 21, 2021 at 17:15
  • $\begingroup$ Asad, note that a volt meter measures voltage DROP across circuit elements, and it is normally voltage drop in a circuit that is more relevant than voltage. Also note that the voltage drop across all parallel circuit elements is the same. $\endgroup$ Commented May 1, 2021 at 0:46

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I think it helps to remember that normal electronics operates in quasi-static limit so charges in the wires are essentially in equilibrium, any current is a slight perturbation on top of otherwise random motion.

Imagine connecting two or more wires into a vertex. Imagine the vertex-facing end of the first wire is at different potential to other wires you just connected. What will happen? The charges from all the wires will rush either towards or away from different-potential wire-end. As they do so, further charges will be repelled, thus very quickly a new equilibrium will be reached where all vertex-facing wire end are at the same electrical potential.

So we have the following 'law': Wires joined together are at the same electrostatic potential.

Now consider parallel and in-series resistors:

enter image description here

Parallel resistors

Imagine there is a voltage drop of $V_1$ across $R_1$. Let $\phi_{W1}=0V$ (we are free to choose the value of electrostatic potential once, since it's the differences between them that matter). Then we know that $\phi_{U1}=\phi_{W1}+V_1=V_1$. But since the three wires are joined at vertex $A$ we also then know that $\phi_U=\phi_{U1}=\phi_{U2}=V_1$. At the same time, three wires are joined at vertex $B$ so $\phi_W=\phi_{W1}=\phi_{W2}=0$. So then the voltage drop across the second resistor is $V_2=\phi_{U2}-\phi_{W2}=V_1$. Voltage drop across both resistors is $V=\phi_U-\phi_W=V_1$.

Series resistors

Let voltage drops across $R_1$ and $R_2$ be $V_1$ and $V_2$, respectively. Choose $\phi_W=0V$. Then we know that $\phi_L=V_2+\phi_W=V_2$. But two wires join at vertex $A$ so $\phi_K=\phi_L=V_2$. The voltage drop across $R_1$ is $V_1$ so $\phi_U=\phi_K+V_1=V_2+V_1$. Thus the voltage drop across both resistors is $\phi_U-\phi_W=V_1+V_2$.

To figure out what happens with current use the same diagrams but remember that charge is essentially incompressible in this setting, so all the current that goes into a vertex, or into a resistor, or any portion of your circuit, must come out of it. So, for example, in case of in-series resistors, at vertex A. If you know current in two out of three wires, the current in the third wire will simply be whatever is needed to ensure that all the current that goes into $A$, leaves it.


You can think of electric potential here as the pressure applied on all the charges. As long as pressure is the same, there is only random motion, but any difference in pressure leads to compensating re-distribution.

The picture I gave above is for electrostatic regime and zero-resistance wires, things become more complex once you depart from these assumptions

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  • $\begingroup$ You're saying that if we take two wires of 2 different P.D(s) and join them together, they will immediately try to reach an equillibrium, correct? But what about the series then? In series there aren't different wires for different circuit component, only 1 wire through which all the load is conected. How does P.D for different component variate when there is only 1 pathway for the charges to flow through? And if possible, please use easy words, I'm a 10th grade student trying to grasp this concept $\endgroup$
    – Asad
    Commented Apr 21, 2021 at 19:29
  • $\begingroup$ Do you understand the reasoning for the parallel resistors? $\endgroup$
    – Cryo
    Commented Apr 21, 2021 at 20:53
  • $\begingroup$ That is my question, I want reasoning behind why does current divide in parallel and remains the same in series, while voltage remains the same in parallel and divides in series. $\endgroup$
    – Asad
    Commented Apr 28, 2021 at 17:22
  • $\begingroup$ To understand you will need to work on it. I cant just transplant my understanding. Let us go back to my question to you. Does the explanation for voltage in parallel resistors make sense to you? Do you understand what electrostatic potential is? Do you understand the statements about equilibrium? Also, you may get your answers faster with a textbook. Are you using one now? $\endgroup$
    – Cryo
    Commented Apr 28, 2021 at 20:15
  • $\begingroup$ As for using a book, I do have a physics book but it only states that in series this happens and in parallel that happens. There are no reasonings given in the book that is why the book just seems useless to me. As for electrostatic potential and electrical equilibrium, yes I do understand them. I feel like I would understand this topic more easily with figures. Sorry for asking you for more however I'm just stuck at this topic. $\endgroup$
    – Asad
    Commented Apr 29, 2021 at 17:36

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